Question

Determine, without graphing, whether the given quadratic function has a maximum value or a minimum value, and then find the value. \(f(x)=4 x^{2}-4 x\)

Short answer

The quadratic function has a minimum value of -1 at \(x = \frac{1}{2}\).

Step-by-step solution

Identify the coefficients

Identify the coefficients from the quadratic function in the form of \(f(x)=ax^{2}+bx+c\).Here, \(a = 4\), \(b = -4\), and \(c = 0\).

Determine the direction of the parabola

The sign of the coefficient \(a\) determines whether the parabola opens upwards or downwards.Since \(a = 4\) is positive, the parabola opens upwards, indicating that the function has a minimum value.

Find the vertex of the parabola

The vertex of a quadratic function \(f(x) = ax^2 + bx + c\) is given by the formula \(x = -\frac{b}{2a}\). Substitute in the values of \(a\) and \(b\):\[ x = -\frac{-4}{2(4)} = \frac{4}{8} = \frac{1}{2} \]

Substitute \(x = \frac{1}{2}\) into the function

Find the minimum value by substituting \(x = \frac{1}{2}\) back into the function \(f(x) = 4x^2 - 4x\):\[ f\left( \frac{1}{2} \right) = 4\left( \frac{1}{2} \right)^{2} - 4\left( \frac{1}{2} \right) = 4\left( \frac{1}{4} \right) - 2 = 1 - 2 = -1 \]

Key concepts

Parabola

The quadratic function is commonly written in the form of \(f(x) = ax^2 + bx + c\). When graphed, it forms a shape called a parabola.The parabola has several important characteristics:

• Direction: The sign of the coefficient \(a\) determines the direction. If \(a > 0\), the parabola opens upwards. If \(a < 0\), the parabola opens downwards.
• Vertex: The turning point of the parabola is called the vertex. It can either be a maximum or minimum point.
• Axis of Symmetry: The parabola is symmetric around the vertical line that passes through the vertex, known as the axis of symmetry.

In the example problem, we see that the coefficient \(a = 4\), which means the parabola opens upwards. This information tells us that the function has a minimum value at its vertex.

Vertex Formula

The vertex of a parabola represented by the quadratic function \(f(x) = ax^2 + bx + c\) can be found using the vertex formula:x = -\(\frac{b}{2a}\).
This formula gives us the x-coordinate of the vertex, where the parabola attains its maximum or minimum value.

• Substitute \(a\) and \(b\) into the formula to find the x-coordinate of the vertex.
• In our example, \(a = 4\) and \(b = -4\). So, \( x = -\frac{-4}{2(4)} = \frac{1}{2} \).

Once we find the x-coordinate, we substitute it back into the original function to find the y-coordinate, which gives us the minimum value since the parabola opens upwards.

Minimum Value

Since the parabola opens upwards (as \(a = 4\)), the function has a minimum value at the vertex.
To find this minimum value, substitute the x-coordinate of the vertex back into the quadratic function.In our case, once we found that the vertex x-coordinate is \(\frac{1}{2}\), we substitute \(x = \frac{1}{2}\) back into \(f(x) = 4x^2 - 4x\):
\[ f\left( \frac{1}{2} \right) = 4\left( \frac{1}{2} \right)^{2} - 4\left( \frac{1}{2} \right) = 4\left( \frac{1}{4} \right) - 2 = 1 - 2 = -1 \]
Hence, the minimum value of the function is -1 at x = \(\frac{1}{2}\).Understanding these concepts helps in solving many problems involving quadratic functions and their graphs.